Integrand size = 52, antiderivative size = 25 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^x}{x+\frac {2 x}{4-\log (\log (5) (4+\log (25)))}} \]
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Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6, 12, 14, 2208, 2209} \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^x (4-\log (\log (5) (4+\log (25))))}{x (6-\log (\log (5) (4+\log (25))))} \]
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Rule 6
Rule 12
Rule 14
Rule 2208
Rule 2209
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{x^2 (-6+\log (4 \log (5)+\log (5) \log (25)))} \, dx \\ & = \frac {\int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{x^2} \, dx}{-6+\log (4 \log (5)+\log (5) \log (25))} \\ & = \frac {\int \left (\frac {4 e^x \left (1-\frac {1}{4} \log (\log (5) (4+\log (25)))\right )}{x^2}-\frac {4 e^x \left (1-\frac {1}{4} \log (\log (5) (4+\log (25)))\right )}{x}\right ) \, dx}{-6+\log (4 \log (5)+\log (5) \log (25))} \\ & = -\frac {(4-\log (\log (5) (4+\log (25)))) \int \frac {e^x}{x^2} \, dx}{6-\log (\log (5) (4+\log (25)))}+\frac {(4-\log (\log (5) (4+\log (25)))) \int \frac {e^x}{x} \, dx}{6-\log (\log (5) (4+\log (25)))} \\ & = \frac {e^x (4-\log (\log (5) (4+\log (25))))}{x (6-\log (\log (5) (4+\log (25))))}+\frac {\text {Ei}(x) (4-\log (\log (5) (4+\log (25))))}{6-\log (\log (5) (4+\log (25)))}-\frac {(4-\log (\log (5) (4+\log (25)))) \int \frac {e^x}{x} \, dx}{6-\log (\log (5) (4+\log (25)))} \\ & = \frac {e^x (4-\log (\log (5) (4+\log (25))))}{x (6-\log (\log (5) (4+\log (25))))} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^x (-4+\log (\log (5) (4+\log (25))))}{x (-6+\log (\log (5) (4+\log (25))))} \]
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Time = 2.76 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32
method | result | size |
risch | \(\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+\ln \left (2+\ln \left (5\right )\right )-4\right ) {\mathrm e}^{x}}{x \left (\ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+\ln \left (2+\ln \left (5\right )\right )-6\right )}\) | \(33\) |
gosper | \(\frac {{\mathrm e}^{x} \left (\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-4\right )}{x \left (\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-6\right )}\) | \(37\) |
norman | \(\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-4\right ) {\mathrm e}^{x}}{\left (\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6\right ) x}\) | \(37\) |
parallelrisch | \(\frac {{\mathrm e}^{x} \ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-4 \,{\mathrm e}^{x}}{\left (\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-6\right ) x}\) | \(41\) |
meijerg | \(\frac {\left (\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-4\right ) \left (\ln \left (x \right )+i \pi -\ln \left (-x \right )-\operatorname {Ei}_{1}\left (-x \right )\right )}{\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-6}-\frac {\left (-\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )+4\right ) \left (\frac {1}{x}+1-\ln \left (x \right )-i \pi -\frac {2+2 x}{2 x}+\frac {{\mathrm e}^{x}}{x}+\ln \left (-x \right )+\operatorname {Ei}_{1}\left (-x \right )\right )}{\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-6}\) | \(125\) |
default | \(-\frac {\ln \left (2\right ) \operatorname {Ei}_{1}\left (-x \right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}-\frac {\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right ) \operatorname {Ei}_{1}\left (-x \right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}+\frac {-\frac {4 \,{\mathrm e}^{x}}{x}-4 \,\operatorname {Ei}_{1}\left (-x \right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}+\frac {4 \,\operatorname {Ei}_{1}\left (-x \right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}-\frac {\ln \left (2\right ) \left (-\frac {{\mathrm e}^{x}}{x}-\operatorname {Ei}_{1}\left (-x \right )\right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}-\frac {\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right ) \left (-\frac {{\mathrm e}^{x}}{x}-\operatorname {Ei}_{1}\left (-x \right )\right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}\) | \(194\) |
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Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^{x} \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right ) - 4 \, e^{x}}{x \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right ) - 6 \, x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {\left (-4 + \log {\left (\log {\left (5 \right )} \right )} + \log {\left (2 \right )} + \log {\left (\log {\left (5 \right )} + 2 \right )}\right ) e^{x}}{- 6 x + x \log {\left (\log {\left (5 \right )} \right )} + x \log {\left (2 \right )} + x \log {\left (\log {\left (5 \right )} + 2 \right )}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 3.92 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {{\rm Ei}\left (x\right ) \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right )}{\log \left (2 \, {\left (\log \left (5\right ) + 2\right )} \log \left (5\right )\right ) - 6} - \frac {\Gamma \left (-1, -x\right ) \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right )}{\log \left (2 \, {\left (\log \left (5\right ) + 2\right )} \log \left (5\right )\right ) - 6} - \frac {4 \, {\rm Ei}\left (x\right )}{\log \left (2 \, {\left (\log \left (5\right ) + 2\right )} \log \left (5\right )\right ) - 6} + \frac {4 \, \Gamma \left (-1, -x\right )}{\log \left (2 \, {\left (\log \left (5\right ) + 2\right )} \log \left (5\right )\right ) - 6} \]
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Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^{x} \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right ) - 4 \, e^{x}}{x \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right ) - 6 \, x} \]
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Time = 9.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {{\mathrm {e}}^x\,\left (\ln \left (\ln \left (625\right )+2\,{\ln \left (5\right )}^2\right )-4\right )}{x\,\left (\ln \left (\ln \left (625\right )+2\,{\ln \left (5\right )}^2\right )-6\right )} \]
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